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I have been trying to studying the construction of Lebesgue integral for a while now. I am following the Princeton Lectures on Analysis and I am stuck at the part where it defines the integral of non-negative functions. I find the definition to be quite clear but I cannot understand the examples given here.

With the above definition of the integral, there are only two possible cases; the supremum is either finite, or infinite. In the first case, when $\int f(x)dx<+\infty$, we shall say that $f$ is Lebesgue integrable or simply integrable.

Clearly, if $E$ is any measurable subset of $\mathbb R^d$, and $f\ge0$, then $f_{\chi_E}$ is also positive, and we define $$\int f(x)dx=\int f(x)\chi_E(x)dx.$$ Simple examples of functions on $\mathbb R^d$ that are integrable (or non-integrable) are given by $$f_a(x)=\begin{cases}|x|^{-a}&\text{ if }|x|\le1,\\0&\text{ if }|x|>1.\end{cases}$$ $$F_a(x)=\frac1{1+|x|^a}, \text{ all }x\in\mathbb R^d.$$ Then $f_a$ is integrable exactly when $a<d$, while $F_a$ is integrable exactly when $a>d$.

How are the values of "a" here in these two examples making the function integrable or not !? and what does the value of "a" has to do with "d" ?

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  • $\begingroup$ The point discussed in the article is not about measurability. All the maps under discussion are measurable. It is about integrability. And it seems that the integrability of those maps is discussed later on in the article. $\endgroup$ – mathcounterexamples.net Jan 13 at 9:45
  • $\begingroup$ Im sorry thats what i meant. How does it make the function integrable or not. I am changing the orginal question. And it doesnt discuss anything about these further down. $\endgroup$ – Rpdp_s Jan 13 at 9:48
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    $\begingroup$ It is written see the discussion following Corollary 1.10 and Exercise 10. What are those about? $\endgroup$ – mathcounterexamples.net Jan 13 at 9:50
  • $\begingroup$ Those are a bit further down after discussing a few more theorems, However, I was worried if I missed something on the way making the example seem hard. Looks like I will have to skip them for now and come back. Thank you for pointing it out. $\endgroup$ – Rpdp_s Jan 13 at 9:54
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Consider a map $f : \mathbb R^d \to \mathbb R$ which is only dependent on the radius $R$ in the $n$-sphere Spherical coordinates. I.e. $f(x)=g(\vert x\vert)$ where $g$ is a real map.

You then have

$$\int_{\mathbb R^d} f(x) \ dx = K_d \int_0^\infty g(R) R^{d-1}\ dR$$ where $K_d$ is a constant that only depends on $d$. This uses an integral by substitution in dimension $d$.

You can use that to look at the integrability of the maps of your original question.

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Likely this is explained further down in your text. Anyway...

The Lebesgue integral of $f_a$ over $\mathbb R^d$ is: $$\int_{\mathbb R^d} f_a\,d\mu = \int_0^\infty \mu\big(\{x\in B_d(1):f_a(x)>y\}\big)\,dy = \int_0^\infty \mu\big(\{x\in B_d(1):|x|^{-a}>y\}\big)\,dy$$ where $B_d(r)$ is the unit ball in $d$ dimensions with radius $r$.

If $a\le 0$ there is no singularity nor infinity involved, and the result is finite.

If $a>0$ then the measure at $y$ is the volume of the $d$-ball $B_d(r)$ with radius $r=y^{-\frac 1a}\le 1$. That also means that $y\ge 1$. The volume of $B_d(r)$ is $C_d r^d$ for some constant $C_d$ that depends only on $d$ (for instance $C_2=\pi$ and $C_3=\frac 43\pi$). So the integral becomes: $$\int_{\mathbb R^d} f_a\,d\mu = \int_1^\infty C_d\cdot(y^{-\frac 1a})^d\,dy = \frac{C_d}{-\frac da+1} y^{-\frac da+1}\Bigg|_1^\infty $$ This is finite iff $-\frac da+1<0 \iff a<d\quad$ (for the case $a>0$).

Therefore $f_a$ is Lebesgue integrable over $\mathbb R^d$ iff $a<d$.

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